Question: Solve for $a$, $ \dfrac{2}{a^2} = \dfrac{9}{3a^2} + \dfrac{3a + 6}{4a^2} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $a^2$ $3a^2$ and $4a^2$ The common denominator is $12a^2$ To get $12a^2$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ \dfrac{2}{a^2} \times \dfrac{12}{12} = \dfrac{24}{12a^2} $ To get $12a^2$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{3a^2} \times \dfrac{4}{4} = \dfrac{36}{12a^2} $ To get $12a^2$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{3a + 6}{4a^2} \times \dfrac{3}{3} = \dfrac{9a + 18}{12a^2} $ This give us: $ \dfrac{24}{12a^2} = \dfrac{36}{12a^2} + \dfrac{9a + 18}{12a^2} $ If we multiply both sides of the equation by $12a^2$ , we get: $ 24 = 36 + 9a + 18$ $ 24 = 9a + 54$ $ -30 = 9a $ $ a = -\dfrac{10}{3}$